\((B,\circ )=(\mathbb {Z}/p^2\mathbb {Z})\rtimes _{\alpha ^{\prime \prime }}\mathbb {Z}/2\mathbb {Z}\) , \(E=\mathbb {Z}/p^2\mathbb {Z}\times \lbrace 0\rbrace \) , \(G=\lbrace 0\rbrace \times \mathbb {Z}/2\mathbb {Z}\) , \((e_1,g_1)+(e_2,g_2)=(e_2,g_1+g_2)\)  and \((e_1,g_1)\circ (e_2,g_2)=(e_1+(-1)^{g_1}e_2,g_1+ g_2)\)  for all \((g_1,e_1),(g_2,e_2)\in B\) ;
By e1p2, the left semi-brace \(B\)  is isomorphic to a semidirect product of the trivial left semi-brace \(E\)  and the trivial skew left brace \(G\)  of size 2. By a long but easy calculation, we have that the isomorphism classes of the semidirect product of left semi-braces \(E\rtimes _{\alpha } G \)  are completely determined by the isomorphism classes of the groups semidirect product \((E\rtimes G,\circ )\) .
Left and right nilpotency of left cancellative left semi-braces
The goal of this section is to study various types of nilpotency, recently introduced in [1]}, for the left semi-braces characterized in Section 2. In particular, we characterize the ones that are right nilpotent and we find sufficient conditions to be right nil or left nil. These results will be applied to partially understand left and right nilpotency of the left semi-braces of size \(pq\)  and \(2p^2\)  obtained in Sections 3 and 4, many of which will allow us to answer in negative to [1]}.
Let \(B\)  be a left cancellative left semi-brace and \(\)  the binary operation on \(B\)  given by
\(ab:=\lambda _{a}(a^-)+a\circ b+\lambda _b(b^-) \) 
for all \(a,b\in B\) . By [1]}, we have that \(ab\in G\) , for all \(a,b\in B\) . Therefore, if \(X\)  and \(Y\)  are subsets of \(B\) , we denote by \(XY\)  the additive subgroup of \((G,+)\)  generated by the elements of the form \(ab\) , where \(a\in X\)  and \(b\in Y\) . In the case \(X=Y:=B\)  one can give two further notions which will be important in this section.
Definition 5.1 Let \(B\)  be a left cancellative left semi-brace. We define the subset \(B^{(n)}\)  and \(B^n\)  by \(B^{(1)}=B^1:=B\)  and
\(B^{(n+1)}:=B^{(n)}B+E\qquad B^{n+1}:=BB^n+E\) 
for all \(n\in \mathbb {N}\) .

In analogy to skew left braces, we say that a left cancellative left semi-brace \(B\)  is right (resp. left) nilpotent if there exist a natural number \(n\)  such that \(B^{(n)}\)  (resp. \(B^n\) ) is equal to the set \(E\) . As a weaker version of the right nilpotency, we call a left cancellative left semi-brace right nil if for every \(b\in B\)  there exist \(n\in \mathbb {N}\)  such that \(b^{(n)}=0\) , where \(b^{(1)}:=b\)  and \(b^{(n+1)}:=b^{(n)}b\)  for all \(b\in B\) . As one can expect, the concept of left nil left semi-brace is defined in a similar way.
As a first result of the section, we characterize left semi-braces of Section 2 that are right nilpotent.
Proposition 5.2 
Let \(B:=G\rtimes _{\alpha } E\)  be a left cancellative left semi-brace as in teostrutt. Then, \(B\)  is right nilpotent if and only if \(\alpha _e=id_G\)  for all \(e\in E\)  and \(G\)  is right nilpotent.

Suppose that \(\alpha _e=id_G\)  for all \(e\in E\)  and \(G\)  is right nilpotent. Then, \(E\)  is an ideal of \(B\)  and therefore by [1]} \(B^{(n)}=G^{(n)}+E\)  for all \(n\in \mathbb {N}\) . Therefore, if \(G\)  is right nilpotent and \(m\)  is a natural number such that \(G^{(m)}=0\) , we have that \(B^{(m)}=E\) .
Conversely, if \(B\)  is right nilpotent we have that \(E\)  is an ideal of \(B\) , hence a normal subgroup of \((B,\circ )\) . This implies that \(\alpha _e=id_G\)  for all \(e\in E\) . Using again the equality \(B^{(n)}=G^{(n)}+E\)  for all \(n\in \mathbb {N}\) , the right nilpotency of \(G\)  also follows.
By the previous result, examples of left cancellative left semi-braces that are not right nilpotent occur in abundance among the ones considered in Section 2. However, a large family of these left semi-braces are right nil, as we can see in the next proposition. At first, we need a technical lemma.
Lemma 5.3 
Let \(B:=G\rtimes _{\alpha } E\)  be a left cancellative left semi-brace as in teostrutt. Then
\((g_1,e_1)(g_2,e_2)=(-g_1+g_1\circ \alpha _{e_1}(g_2)-g_2,0)\) 
for all \(g_1,g_2\in G\) , \(e_1,e_2\in E\) .

It follows by an easy calculation.
Proposition 5.4 
Let \(B:=G\rtimes _{\alpha } E\)  be a left cancellative left semi-brace as in teostrutt. Moreover, suppose that \(G\)  is a trivial skew left brace. Then, \(b^{(3)}=0\)  for all \(b\in B\) .

By nilel, \((g_1,e_1)^{(2)}=(\alpha _{e_1}(g_1)-g_1,0) \)  and
\((g_1,e_1)^{(3)}=(\alpha _{e_1}(g_1)-g_1,0)(g_1,e_1)=(\alpha _0(g_1)-g_1,0)=(0,0) \) 
for all \(g_1\in G\) , \(e_1\in E\) , hence the thesis.
As implicitly showed in [1]}, every right nilpotent left semi-brace is right nil. On the other hand, rnilp1 and rnilp2 state that, using the left cancellative left semi-braces characterized in Section 2, we can construct several examples of finite left cancellative left semi-braces that are right nil but not right nilpotent: as concrete examples, we consider below the left semi-braces obtained in Sections 3 and 4. Therefore, a large part of these left semi-braces answers in negative to [1]}.
Examples 1
1)
Let \(B\)  be a left cancellative left semi-brace of size \(pq\)  (where \(p,q\)  are prime numbers with \(p>q\) ) and with \(|E|>1\) . Then, if \(|E|=p\)  or if \((B,\circ )\)  is abelian, \(B\)  is right nilpotent by [1]}. If \(|E|=q\)  and \((B,\circ )\)  is not abelian, then \(B\)  is isomorphic to the semidirect product of left semi-braces \(G\rtimes _{\alpha } E\) , therefore by rnilp1 and rnilp2 \(B\)  is a right nil left semi-brace which is not right nilpotent.
2)
Let \(B\)  be a left cancellative left semi-brace of size \(2p^2\) . If \(|E|=p^2\)  or if \((B,\circ )\)  is abelian, by [1]} it is right nilpotent. While, if \(|E|=2\) , \((B,\circ )\)  is not abelian and \(G\)  is a trivial skew left brace of size \(p^2\) , by rnilp1 and rnilp2 \(B\)  is a left cancellative left semi-brace right nil but not right nilpotent.


In the rest of this section we focus on left nilpotency. In [1]} a family of left nilpotent left semi-braces was exhibited. In particular, they provided left nilpotent left semi-braces having nilpotent multiplicative group and with \(E\)  contained in the centralizer of \(G\)  or such that is a normal subgroup of \((B,\circ )\) . Below, we show sufficient conditions for the left semi-braces of teostrutt which ensure left nilpotency.
Proposition 5.5 
Let \(B:=G\rtimes _{\alpha } E\)  be a left cancellative left semi-brace as in teostrutt and suppose that \((G,+,\circ )\)  is a skew left brace with cyclic additive group and abelian multiplicative group such that the number \(s:=|G/Soc(G)|\)  is a nilpotent element of the ring \((G,+,*)\)  (where \(*\)  is the usual scalar multiplication). Then, \(B\)  is a left nilpotent left cancellative semi-brace.

Since \((G,+,\circ )\)  has cyclic additive group and abelian multiplicative group, by [10]} there exist \(d\in \mathbb {N}\setminus \lbrace 1\rbrace \)  such that \(d=|Soc(G)|\)  and \(g_1\circ g_2=g_1+g_2+g_1 g_2 d\)  for all \(g_1,g_2\in G\) . By [11]}, we have that \(d\)  is a nilpotent element of the ring \((G,+,*)\) . Now, since \(\lambda _{g_1}(g)=(1+dg_1)g\)  for all \(g,g_1\in G\)  and by [10]} for all \(e\in E\)  there exist \(k_e\in \mathbb {N}\)  such that \(\alpha _e(g)=(1+k_e s) g\)  for all \(g\in G\) , by nilel, we have that
\((g_1,e_1)(g_2,e_2)=([(1+dg_1)(1+k_{e_1}s)-1]g_2,0)=((k_{e_1} s+dg_1+dk_{e_1} g_1 s)g_2,0)\) 
for all \(g_1,g_2\in G\) , \(e_1,e_2\in E\) . Now, if \(\lbrace p_1,...,p_z\rbrace \)  is the set of prime numbers dividing \(|G|\) , by a standard induction and using the nilpotency of \(d\)  and \(s\) , one can show that \(B^n\subseteq (p_1...p_z)^{n-1} G+E \)  for all \(n\in \mathbb {N}\)  and this clearly implies that \(B\)  is left nilpotent.
leftnilpo can be used to construct left nilpotent left cancellative semi-braces that are different from the ones obtained in [1]}, as we see in the following example.
Example 1 Let \(G:=(\mathbb {Z}/p^2\mathbb {Z},+,\circ )\)  be the skew left braces given by \(a\circ b:=a+b+pab\)  for all \(a,b\in \mathbb {Z}/p^2\mathbb {Z}\)  and \(E\)  the trivial left semi-brace on \(\mathbb {Z}/p\mathbb {Z}\) . Let \(B\)  be the left semi-brace given by \(B:=G\rtimes _{\alpha } E\)  where \(\alpha :E\longrightarrow Aut(G,+,\circ )\)  is given by \(\alpha _e(g):=(1+pe)g\)  for all \(e\in E\) , \(g\in G\) . By leftnilpo \(B\)  is a left nilpotent left cancellative semi-brace. On the other hand, \(E\)  is not contained in the centralizer of \(G\)  and it is not a normal subgroup of \((B,\circ )\)  hence it can not be constructed by [1]}.

However, there exist examples of left cancellative left semi-braces provided by teostrutt that are not left nilpotent, as we can note by the following example.
Example 2 
Let \(B\)  be a left cancellative left semi-braces of size \(pq\)  (where \(p,q\)  are prime numbers, \(p>q\) ). Then, if \((B,\circ )\)  is abelian or \(|E|=p\) , since \(G\)  is a trivial skew left brace of size \(q\) , by [1]} we have that \(B\)  is left nilpotent. If \((B,\circ )\)  is not abelian and \(|E|=q\) , then \(B\)  is a semidirect product of left semi-braces as in teostrutt. By a standard calculation, we obtain that \(B^2=B\) , hence \(B\)  is not left nilpotent.

We conclude the paper remarking that in contrast to the right-case, the left semi-braces of size \(pq\)  are left nil if and only if they are left nilpotent, therefore Question 2 of [1]}, where the authors ask if a finite left nil is left nilpotent, remains open.
